question_answer

Show that area of the parallelogram whose diagonals are given by \[\vec{a}\] and \[\vec{b}\] is \[\frac{|\vec{a}\,\,\times \vec{b}|}{2}.\] Also, find the area of the parallelogram whose diagonals are \[2\hat{i}-\hat{j}+\hat{k}\] and \[\hat{i}+3\hat{j}-\hat{k}.\]

Answer:

Let ABCD be a parallelogram such that \[\overrightarrow{AC}=\vec{a}\] and \[\overrightarrow{BD}=\vec{b}.\] Also, let \[\overrightarrow{AB}=\vec{p}\] and \[\overrightarrow{AD}=\vec{q},\] then \[\overrightarrow{BC}=\vec{q}\] and \[\overrightarrow{DC}=\vec{p}.\] Now, by triangle law of addition, we get             \[\overrightarrow{AC}=\vec{p}+\vec{q}\]                                    \[\Rightarrow \]   \[\vec{a}=\vec{p}+\vec{q}\]                              ?(i) Similarly, \[\overrightarrow{BD}=-\,\vec{p}+\vec{q}\] \[\Rightarrow \]   \[\vec{b}=-\,\vec{p}+\vec{q}\]                           ?(ii) On adding Eqs. (ii) and (i), we get             \[\vec{a}+\vec{b}=2\vec{q}\] \[\Rightarrow \] \[q=\frac{1}{2}(\vec{a}+\vec{b})\] ?(iii) On subtracting Eq. (ii) from Eq. (i), we get             \[\vec{a}-\vec{b}=2\vec{p}\] \[\Rightarrow \]   \[\vec{p}=\frac{1}{2}(\vec{a}\,-\vec{b})\] Now, \[\vec{p}\times \vec{q}=\frac{1}{4}\{(\vec{a}-\vec{b})\times (\vec{a}+\vec{b})\}\]                         [using Eqs. (iii) and (iv)]             \[=\frac{1}{4}\{\vec{a}\times \vec{a}+\vec{a}\times \vec{b}-\vec{b}\times \vec{a}-\vec{b}\times \vec{b}\}\]             \[=\frac{1}{4}\{\vec{a}\,\times \vec{b}+\vec{a}\times \vec{b}\}\] \[\left[ \begin{align}   & \because \,\,\,\vec{a}\times \vec{a}=\vec{b}\times \vec{b}=\vec{0} \\  & \text{and}\,\,\vec{a}\,\times \vec{b}=-\vec{b}\times \vec{a} \\ \end{align} \right]\]             \[=\frac{1}{4}\times 2(\vec{a}\times \vec{b})=\frac{1}{2}(\vec{a}\times \vec{b})\] So, area of a parallelogram \[ABCD=\frac{1}{2}|\vec{a}\times \vec{b}|\] Now, area of a parallelogram whose diagonals are \[2\hat{i}-\hat{j}+\hat{k}\] and \[\hat{i}+3\hat{j}-\hat{k}\]             \[=\frac{1}{2}|(2\hat{i}-\hat{j}+\hat{k})\times (\hat{i}+3\hat{j}-\hat{k})|\]                         \[=\frac{1}{2}|\hat{i}(1-3)-\hat{j}(-\,2-1)+\hat{k}(6+1)|\] \[=\frac{1}{2}|2\hat{i}+3\hat{j}+7\hat{k}|\,\,\,=\frac{1}{2}\sqrt{{{(-\,2)}^{2}}+{{(3)}^{2}}+{{(7)}^{2}}}\]             \[=\frac{1}{2}\sqrt{4+9+49}\]             \[=\frac{1}{2}\sqrt{62}\,\,\text{sq}\,\,\text{units}\]